The standard calculation
4 marks · NormalA table shows the results from a titration between lithium hydroxide ( ) and hydrochloric acid ( ).
| Volume of LiOH (cm³) | Concentration of LiOH (mol/dm³) | Volume of HCl (cm³) | Concentration of HCl (mol/dm³) |
|---|---|---|---|
| 25.0 | 0.100 | 20.0 | ? |
Write the balanced chemical equation for this reaction.
Calculate the number of moles of LiOH used.
Using the chemical equation, state the number of moles of HCl that reacted.
Calculate the concentration of the HCl solution.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
Show answer
(a) \ce{LiOH (aq) + HCl (aq) -> LiCl (aq) + H2O (l)}
(b) 0.00250 mol
(c) 0.00250 mol
(d) 0.125 mol/dm³
Common mistakes: not converting from cm³ to dm³; incorrectly applying the stoichiometric ratio
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Writes correctly balanced equation for the neutralisation.
M1 (b) · 1 mark · Uses concentration × volume (in dm³) for LiOH: 0.100 × 0.0250 = 0.00250 mol.
M2 (c) · 1 mark · Uses equation ratio (1:1) to deduce moles of HCl reacted = 0.00250 mol. (follow-through)
A1 (d) · 1 mark · Works out concentration of HCl: 0.00250 ÷ 0.0200 = 0.125 mol/dm³. (follow-through)
Worked solution
(a) The balanced equation is: \ce{LiOH (aq) + HCl (aq) -> LiCl (aq) + H2O (l)}
(b) Volume of LiOH = 25.0 cm³ = 0.0250 dm³ Moles of LiOH = 0.100 × 0.0250 = 0.00250 mol
(c) Ratio is 1:1, so moles of HCl = 0.00250 mol
(d) Volume of HCl = 20.0 cm³ = 0.0200 dm³ Concentration HCl = 0.00250 ÷ 0.0200 = 0.125 mol/dm³